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Let $\phi\in X'$ , where X is a Hilbert space and $X'$ its dual. Then I want to check that $\ker\phi$ is a closed subspace of $X$ of codimension 1 ($\phi \ne 0$).

So to see the closeness we pick a sequence $(x_n) \in \ker \phi$ that converges to $x$, then, since $\phi$ is a bounded functional it is continuous, thus $\phi((x_n)) \to \phi(x)=0$ therefore $\ker \phi$ is closed.

The thing is that I don't know how to attached the socond part, because I think I need to prove that the space $X/ \ker \phi=\{x+\ker \phi : x \in X\}$ has dimension 1.

Can someone help me to prove this assertion please?

Thanks a lot in advance.

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    $\begingroup$ Note that the image of $\phi$ is $\mathbb R$ or $\mathbb C$ (or whatever field your Hilbert space is over). $\endgroup$
    – SamM
    Nov 11, 2015 at 18:02
  • $\begingroup$ Can you elaborate more in this please? Thanks :) $\endgroup$
    – user162343
    Nov 11, 2015 at 18:08
  • $\begingroup$ The dual space consists of all bounded linear functionals, which are linear maps from a normed (Banach, Hilbert, etc.) space $X$ into the field $\mathbb K$, which is usually $\mathbb R$ or $\mathbb C$. Thus the image of $\phi$ is isomorphic (as a vector space) to $\mathbb K$. $\endgroup$
    – SamM
    Nov 11, 2015 at 18:11
  • $\begingroup$ Rigth :), but how does that shows that the kernel has codimension 1? $\endgroup$
    – user162343
    Nov 11, 2015 at 18:14
  • $\begingroup$ Don't forget you need $\phi \ne 0$... $\endgroup$
    – peter a g
    Nov 11, 2015 at 18:14

2 Answers 2

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This is a purely algebraic result that you can prove without using the inner product or any other additional structure on your vector space:

Proposition. Let $V$ be a vector space over some field $\mathbb{F}$, and let $\phi\colon V\to\mathbb{F}$ be a nontrivial linear functional. Then $\ker(\phi)$ has codimension 1.

The proof will consist of the following three steps:

  1. Note that, by the first isomorphism theorem, $V/\ker(\phi)\cong\operatorname{im}(\phi)$ (as $\mathbb{F}$-vector spaces).

  2. Note that $\operatorname{im}(\phi)$ is a subspace of $\mathbb{F}$, and that the only subspaces of $\mathbb{F}$ are $\{0\}$ and $\mathbb{F}$ itself. Since $\phi$ is nontrivial, the only possibility is $\operatorname{im}(\phi)=\mathbb{F}$.

  3. Use the fact that isomorphic spaces have the same dimension to conclude that $\operatorname{codim}_\mathbb{F}(\ker(\phi))\overset{\mathrm{def}}{=}\dim_\mathbb{F}(V/\ker(\phi))=\dim_\mathbb{F}(\mathbb{F})=1$.

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Let $z\not\in \ker\phi$, the observe that for each $x\in X$, $$x= \frac{\phi(x)}{\phi(z)} z - \frac{\phi(x)}{\phi(z)} z + x\\ = \frac{\phi(x)}{\phi(z)} z + \bigg[x - \frac{\phi(x)}{\phi(z)} z \bigg]$$ and $\bigg[x - \frac{\phi(x)}{\phi(z)} z \bigg] \in ker \phi$. This means that $X$ can be written as the direct sum of its two subspaces that is $$X = \text{span} \{z \}\oplus \ker \phi.$$ Therefore the $\ker \phi$ has codim $1$.

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    $\begingroup$ Thank you , let me check it , if I have further questions , can I let you know ? $\endgroup$
    – user162343
    Nov 11, 2015 at 19:43
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    $\begingroup$ @user162343 Sure! Glad to help. $\endgroup$
    – Xiao
    Nov 11, 2015 at 19:44
  • $\begingroup$ @Xiao Correct me if I'm wrong, but this proof works for any linear functional $\phi$ on a vector space $X$, right? $\endgroup$
    – zxmkn
    Jan 27, 2019 at 21:00
  • $\begingroup$ @zxmkn Yes, you are correct $\endgroup$
    – Xiao
    Jan 28, 2019 at 20:58

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